Matemáticas Cálculo de derivadas PROBLEMAS ... - e … resueltos de... · Tomamos ln antes de...

Matemáticas Cálculo de derivadas

IES Fernando de Herrera – Prof. R. Mohigefer Página 1 de 17 http://www.e-matematicas.es

PROBLEMAS RESUELTOS DE CÁLCULO DE DERIVADAS

1) Calcular las derivadas de:

a) 52

( )cos

xf x

x

x

xxxxxf

2

54

cos

sen 2cos10)('

b) 3

( ) ln 72

g x x

Simplificamos antes de derivar, aplicando propiedades de logaritmos

neperianos: g(x) = x7ln2

1

2

3 = x7ln

4

3

xxg

7

7

4

3)(' =

x4

3

c) 3 5

( )2

xeh x

Como 53

2

1)( xexh 533

2

1)(' xexh = 53

2

3 xe

2) Halle f ’(2), g ’(4) y h ’(0) para las funciones definidas de la siguiente forma (L

designa logaritmo neperiano):

2

2

16( )f x x

x ; 2 3( ) ( 9)g x x ; 2( ) L( 1)h x x .

Simplemente, aplicando las reglas de derivación, se obtiene:

4

32'( ) 2

xf x x

x =

3

322x

x f ’(2) =

324

8 = 4–4 = 0

2 2'( ) 3( 9) 2g x x x = 2 26 ( 9)x x g’(4) = 24(16+9)2 = 15.000

2

2'( )

1

xh x

x

h’(0) =

0

1 = 0

3) Derivar y simplificar:

a) 2 2( .

3 1( ) 5 )

xf x x x

x

f '(x) = )25)(5(2)13(3 2

2xxx

x

xx

= )251025(2

1 322

2xxxx

x =

= )25152(21 23

2xxx

x = xxx

x50304

1 23

2 =

= 2

345 150304

x

xxx

b) 2( ) ( 1) ln . g x x x

g '(x) = 2x ln x + x

x 12 =

x

xxx 1ln2 22



c) 5 .( ) 2 xh x

h '(x) = 5·25x

ln 2

d) 3 2 3( ) ( 6 ) ( 1 .)i x x x x

i '(x) = (3x2 – 6)(x

2 + 1)

3 + (x

3 – 6x) 3 · 2x (x

2 + 1)

2 =

= (x2 + 1)

2 ((3x

2 – 6) (x

2 + 1) + 6x(x

3 – 6x)) =

= (x2 + 1)

2 (3x

4 + 3x

2 – 6x

2 – 6 + 6x

4 – 36x

2) = (x

2 + 1)

2 (9x

4 – 39x

2 – 6)

e) 2 1.( ) ( 1) xj x x e

j '(x) = e2x + 1

+ (x + 1) 2 e2x + 1

= e2x + 1

(1 + 2x + 2) = e2x + 1

(2x + 3)

f) 2( ) 3 cos3k x x x .

k '(x) = 3 cos 3x2 – 3x·6x sen 3x

2 = 3 cos 3x

2 – 18x

2 sen 3x

2

Nota: La expresión simplificada final siempre puede resultar subjetiva, y debe

entenderse como una expresión cómoda para operar y para volver a derivar si es

preciso. Por ejemplo, en el d y el f se podría extraer 3 factor común.

4) Calcule las derivadas de las siguientes funciones:

a) f(x) =

2

2

2 5 1 2

3

x x

x

f '(x) = 4

2 2)21(2

3

5

3

522

x

xxxx

=

4

22 422

9

)52(10

x

xxxx

=

= 4

2 22

9

2050

x

xxx

=

4

)22(

9

2050

x

xxx

=

3

22

9

2050

x

xx

=

= 33

34

9

1818

9

2050

x

x

x

xx

=

3

32

9

18182050

x

xxx

b) g(x) = (3x + 2)2 ln(1 + x

2)

g '(x) = 2(3x + 2)3ln(1 + x2) + (3x + 2)

2

21

2

x

x

=

= (18x + 12)ln(1 + x2) +

2

2

1

)23(2

x

xx

c) h(x) = 25x

+ 2

1

x

h'(x) = 3

5 22ln2·5

x

x


a) 2 2( .3 1

( ) 5 )x

f x x xx

f '(x) = )25)(5(2)13(3 2

2xxx

x

xx

= )410)(5(

133 2

2xxx

x

xx

=



= )4102050(1 322

2xxxx

x =

2

542 430501

x

xxx =

= 2

245 150304

x

xxx

b) 2( ) ( 1) ln . g x x x

g'(x) = 2x ln x + x

x 12

c) 3 .( ) 2 xh x

h'(x) = 3·23x

ln 2

d) 3 2 3( ) ( 6 ) ( 1 .)i x x x x

i'(x) =(3x2 – 6)(x

2 + 1)

3 + (x

3 – 6x)3(x

2 + 1)

22x =

= (3x2 – 6)(x

2 + 1)

3 + (6x

4 – 36x

2)(x

2 + 1)

2 =

= (x2 + 1)

2[(3x

2 – 6)(x

2 + 1) + 6x

4 – 36x

2] =

= (x2 + 1)

2(3x

4 + 3x

2 – 6x

2 – 6 + 6x

4 – 36x

2) =

= (x2 + 1)

2(9x

4 – 39x

2 – 6)

6) Calcular las derivadas de:

a) sen

1 cos

xy

x

y ' =

2)cos1(

)sen (sen )cos1(cos

x

xxxx

=

= 2

22

)cos1(

sen coscos

x

xxx

=

2)cos1(

1cos

x

x

=

2)cos1(

)cos1(

x

x

=

xcos1

1

b) y = arctg(e–2x

) y ' = 22

2

)(1

2x

x

e

e

=

x

x

e

e4

2

1

2

c) 3sen 3y x y ' = 3 (sen2 3x) (3 cos 3x) = 9 sen

2 3x cos 3x

d) 3( 2)

ln2 1

xy

x

= ln (x – 2)

3 – ln 12 x = 3 ln(x – 2) –

2

1ln(2x – 1)

y ' = 12

2

2

1

2

13

xx =

12

1

1

1

2

3

xx =

12

1

2

3

xx

e) 3 3xy x e y ' = 3x2 e

–3x + x

3 (–3) e

–3x = 3x

2 e

–3x – 3x

3 e

–3x =

= 3x2 e

–3x (1 – x)

7) Derivar y simplificar: 2arctg 3y x ; 2 1

3

xx ey

; 2

3( 2)

ln 3

xy

x

; 22cos 4y x

2arctg 3y x y ' = 22 )3(1

6

x

x

= 491

6

x

x

2 1

3

xx ey

= xex 12

3

1 y ' = ))1(2(

3

1 121 xx exxe = 3

)2(1 xxe x

2

3( 2)

ln 3

xy

x

=

3

)2(ln

3

1 2

x

x = )3ln()2ln(

3

1 2 xx =

= )3ln()2ln(23

1 xx y ' =

3

1

2

12

3

1

xx =



= )3(3

1

)2(3

2

xx

22cos 4y x y ' = 2·2 (cos 4x) (–sen 4x) 4 = – 16 sen 4x cos 4x =

= – 8 · 2 sen 4x cos 4x = – 8 sen (2·4x) = – 8 sen 8x


a) y = e2x

tg x y' = 2e2x

tg x + e2x

(1 + tg2 x) = e

2x (tg

2 x + 2 tg x + 1)

b) 2

32

ln 4

xy

x

= )4ln(ln2

3

1 2 xx y' =

4

22

3

12x

x

x

c) y = 2cos3 3x y' = 2·3 cos

2 3x (– sen 3x) · 3 = – 18 sen 3x cos

2 3x

d) y = arcsen x3 y' =

23

2

)(1

3

x

x

=

6

2

1

3

x

x

9) Derivar las siguientes funciones, simplificando los resultados:

a) cos32 xy e 3)3sen (2' 3cos xey x = – 6 ecos 3x

sen 3x

b) arctg 2y x

221

22

2

'x

xy

=

x

x

21

2

1

=

x

x

x

x

21

2

2

2

1

=

x

x

x

21

2

2

=

)21(2

2

xx

x

=

xx

x

24

22

c) 2

3(2 3)

ln 3

xy

x

=

3

)32(ln

3

1 2

x

x = )3ln()32ln(

3

1 2 xx =

= )3ln()32ln(23

1 xx . Derivando:

3

1

32

22

3

1'

xxy =

3

1

32

4

3

1

xx

d) 3 tg 4y x x x

xxy4cos

434 tg3'

2 =

x

xx

4cos

124 tg3

2

e) 22 sen 3y x

y' = 4 (sen 3x cos 3x) 3 = 6 · 2 sen 3x cos 3x = 6 sen 2·3x = 6 sen 6x


a) sen 32 xy xe

y ' = 2esen 3x

+ 2xesen 3x

(cos 3x)3 = 2esen 3x

(1 + 3x cos 3x)

b) 2

3(4 3)

ln 1

xy

x

Simplificando la expresión antes de derivar, aplicando propiedades de

logaritmos:



y = )]1ln()34[ln(3

1 2 xx = )]1ln()34ln(2[3

1 xx

y ' =

1

1

34

42

3

1

xx =

1

1

34

8

3

1

xx

c) cos23 xy x

Tomamos ln antes de derivar:

ln y = ln 3xcos 2x

= ln 3 + ln xcos 2x

= ln 3 + cos 2x ln x

Derivando miembro a miembro:

xxxx

y

y 1)2(cosln2sen 20

' =

x

xxx

2cosln2sen 2 =

= x

xxxx 2cosln2sen 2

y ' = x

xxxxx x 2cosln2sen 2

3 2cos = )ln2sen 22(cos3 12cos xxxxx x

d) 22 sen 3y x

y ' = 2·2 sen 3x (cos 3x) 3 = 6 · 2 sen 3x cos 3x = 6 sen 2·3x = 6 sen 6x


a) 2cos 3 xy xe

y ' = 3ecos x²

+ 3xecos x²

2x(–sen x2) = 3e

cos x²(1 – 2x

2 sen x

2)

b) 2

52

4 3ln

( 1)

xy

x

Simplificando la expresión antes de derivar, aplicando propiedades de

logaritmos:

y = 2 21[ln(4 3) ln( 1) ]

5x x = 21

[ln(4 3) 2ln( 1)]5

x x

y ' = 2

1 8 12

5 4 3 1

x

x x

=

2

1 8 2

5 4 3 1

x

x x

c) y = arctg 3x

y ' =

2

3

2 3

1 3

x

x

=

3

2 3

1 3

x

x =

3

2 3 (1 3 )x x

d) 23 cos 5y x

y ' = 3·2 cos 5x (–5 sen 5x) = – 15·2 sen 5x cos 5x =

= – 15 sen 2·5x = – 15 sen 10x


f(x) = 22x x

x

; g(x) = (x

2 + 1)

2 · ln(e

3x + 4)



f '(x) = 2

2 1)·2()22ln2(

x

xxx xx =

2

22 222ln2

x

xxx xx =

2

2 22ln2

x

xx xx

g'(x) = 2(x2 + 1)2x ln(e

3x + 4) + (x

2 + 1)

2

4

33

3

x

x

e

e =

= 4

)1(3)4ln()44(

3

22333

x

xx

e

xeexx


a) y = 2(7x3 – 3x)

6

y ' = 2·6(7x3 – 3x)

5(21x

2 – 3) = 12(21x

2 – 3) (7x

3 – 3x)

5 =

= (252x2 – 36) (7x

3 – 3x)

5

b) y = 23 12

1

x

x

y ' = 2

2

)1(

1)·123()1(6

x

xxx =

2

22

)1(

12366

x

xxx =

2

2

)1(

1263

x

xx

c) y = 22 1x

y ' = 122

4

2 x

x =

12

2

2 x

x

d) y = (x + 1)e2x + 1

y ' = 1· e2x + 1

+ (x + 1)2 e2x + 1

= e2x + 1

[1 + 2(x + 1)] = e2x + 1

(1 + 2x + 2) =

= e2x + 1

(2x + 3)


a) y = 2(7x2 – 3x)

5

y ' = 2·5(7x2 – 3x)

4(14x – 3) = 10(14x – 3) (7x

2 – 3x)

4

b) y = 4

1

3 2

x

x

y ' = 24

34

)23(

)1(1223

x

xxx =

24

344

)23(

121223

x

xxx =

24

34

)23(

2129

x

xx

c) y = sen 2x

y ' = xx

2cos22

2 =

x

x

2

2cos

d) y = e2x + 1

ln 3x

y ' = 2e2x + 1

ln 3x + e2x + 1

x3

3 =

xxe x 1

3ln212 = x

xxe x 3ln2112



15) Derivar: 3( 2)

ln2 1

xy

x

; y = 2 4(3 1)xe x (1 punto)

3( 2)

ln2 1

xy

x

La simplificamos antes de proceder a derivar:

y = ln (x – 2)3 – ln 12 x = 3 ln (x – 2) –

2

1ln(2x – 1). De donde:

y ' = 12

2

2

1

2

13

xx =

12

1

2

3

xx =

)12)(2(

)2()12(3

xx

xx =

)12)(2(

236

xx

xx =

= )12)(2(

15

xx

x =

242

152

xxx

x =

252

152

xx

x

Cualquiera de las tres expresiones recuadradas valdría como final.

y = 2 4(3 1)xe x y ' = 3242 12)13(2 xexe xx = ]12)13(2[ 342 xxe x =

= )1226( 342 xxe x = )2126( 342 xxe x

16) Derivar y simplificar: 3cos 2y x ; ln 1y x x ; arcsen 1y x ; 3 xy e

a) 3cos 2y x y ' = 3 cos22x (–2sen 2x) = – 6 sen 2x cos

22x

b) ln 1y x x = )1ln(2

1xx y ' =

1

1

2

1)1ln(

2

1

xxx =

=

1)1ln(

2

1

x

xx

c) arcsen 1y x y ' =2)1(1

1

x=

)12(1

1

2 xx =

121

1

2 xx=

= 22

1

xx

d) 3 xy e y ' = xe

x

3

2

13 =

x

e x

2

3 3


a) f(x) = 2

3 1

( 2)

x

x

f '(x) =

4

2

)2(

)2(2)13()2(3

x

xxx =

= 4)2(

]2)13()2(3)[2(

x

xxx =

3)2(

2663

x

xx =

3)2(

83

x

x

b) g(x) = 3

3

3

x

x

g(x) = 3 3

3

x

x =

3 22

3 22

33

3

3

3

x

x

x

x =

3 33

3 22

3

33

x

xx =

x

xx

3

33 3 22

= 3 223 x = 3 29x

g '(x) = 3 22 )9( 3

18

x

x =

3 443

6

x

x =

3 3 3

6

xx

x =

3 3

2

x



De otra forma: g(x) = 3 3

3

x

x = 3

11

)3(

x = 3

2

)3( x g'(x) = 3·)3(3

2 13

2

x = 3

1

)3(2

x =

= 3/1)3(

2

x =

3 3

2

x

c) h(x) = 2xe5x²

h'(x) = 2 e5x²

+ 2x 10x e5x²

= 2 e5x²

(1 + 10x2)

d) j(x) = 3

52

(4 1)ln

3 1

x

x

j(x) = 52

3

13

)14( ln

x

x =

13

)14( ln

5

12

3

x

x = )]13ln()14( [ln

5

1 23 xx =

= )]13ln()14( ln3[5

1 2 xx j'(x) =

13

6

14

12

5

12x

x

x

18) Calcular y simplificar las derivadas de las siguientes funciones:

a) f(x) = 22(3 1)

3 1

x

x

f '(x) = 2

2

)13(

3)13(2)13(3)13(4

x

xxx =

2)13(

)]13(6)13(12)[13(

x

xxx =

= 2)13(

)6181236)(13(

x

xxx =

2)13(

)1818)(13(

x

xx =

2)13(

)1)(13(18

x

xx =

= 2

2

)13(

)133(18

x

xxx =

2

2

)13(

)123(18

x

xx

b) g(x) = (x2 – x + 1) e

5x

g'(x) = (2x – 1) e5x

+ 5(x2 – x + 1) e

5x = e

5x(2x – 1 + 5x

2 – 5x + 5) =

= e5x

(5x2 – 3x + 4)

c) j(x) = 3

54

(5 3)ln

2

x

x

Simplificamos antes de derivar, aplicando propiedades de logaritmos:

j(x) =

5

4

3

2

)35(ln

x

x =

4

3

2

)35(ln

5

1

x

x = )]2ln()35[ln(

5

1 43 xx =

= ))]ln()2(ln()35ln(3[5

1 4xx = )]ln(4)2ln()35ln(3[5

1xx

Y derivamos:

j '(x) =

xx

140

35

53

5

1 =

xx

4

5

1

35

5·3

5

1

=

xx 5

4

35

3

d) h(x) = 233 5 1x

h'(x) = 3 22 )15(3

103

x

x =

3 22 )15(

10

x

x =

3 2

3 2

3 22 15

15

)15(

10

x

x

x

x =

3 32

3 2

)15(

15 10

x

xx=



= 15

15 102

3 2

x

xx


a) y = e3x

sen x

y' = 3e3x

sen x + e3x

cos x = e3x

(3sen x + cos x)

b) y = cos2 4x

y' = 2(cos 4x)(–4 sen 4x) = –8 sen 4x cos 4x = –4·2 sen 4x cos 4x =

= –4 sen (2·4x) = –4 sen 8x

c) arctg 6x2

y' = 22 )6(1

12

x

x

=

4361

12

x

x

d) y = 3

54

(5 3)ln

2

x

x

Simplificamos antes de derivar, aplicando propiedades de logaritmos:

j(x) =

5

4

3

2

)35(ln

x

x =

4

3

2

)35(ln

5

1

x

x = )]2ln()35[ln(

5

1 43 xx =

= ))]ln()2(ln()35ln(3[5

1 4xx = )]ln(4)2ln()35ln(3[5

1xx

Y derivamos:

j '(x) =

xx

140

35

53

5

1 =

xx

4

5

1

35

5·3

5

1

=

xx 5

4

35

3

=

)35(5

512

xx

x

20) Calcular y simplificar las derivadas de las siguientes funciones: (1,6 puntos)

a) y = 23x

cos x

y' = 3·23x

(ln 2)cos x – 23x

sen x = 23x

(3cos x ln 2 – sen x)

b) y = cos 4x2

y' = –8x sen 4x2

c) y = 3 5 2x

y' = 3 2)25( 3

5

x =

)25(3

25 5 3

x

x

d) y = 2

2log (3 5)x

y' = 2ln

1

53

62

x

x

21) Derivar y simplificar las siguientes funciones:

a) y = 2x cos (5x

3 – 2)

y ' = 2x cos (5x

3 – 2) ln 2 – 2

x sen (5x

3 – 2) 15x

2 =

= 2x [cos (5x

3 – 2) ln 2 – 15x

2 sen (5x

3 – 2)]



b) 3

54

6ln

(5 2 )

xy

x

3

54

6ln

(5 2 )

xy

x

=

3

4

1 6ln

5 (5 2 )

x

x =

3 41[ln 6 ln(5 2 ) ]

5x x =

= 1

5[ln 6 + ln x

3 – 4ln(5 – 2x)] =

1

5[ln 6 + 3ln x – 4ln(5 – 2x)]

y ' = 1 3 2

0 45 5 2x x

= 1 3 8

5 5 2x x

c) y = arctg e3x

y ' = 3

3 2

3

1 ( )

x

x

e

e =

3

6

3

1

x

x

e

e

d) y = log (2x4 + 1)

y ' = 3

4

8 1

2 1 ln10

x

x

22) Derivar y simplificar las siguientes funciones:

a) y = 342 4x

y ' = 2

3 34

122

4 (4 )

x

x =

2

3 94

6

4

x

x =

2

2 3 84

6

(2 )

x

x x =

2

2 64

6

2

x

x x =

24

6

2 2 x =

4

3

4x

b) y = tg (5x3 + 1) y ' =

2

2 3

15

cos (5 1)

x

x

c) y = ex(4x

3 + 2)

3

y ' = ex(4x

3 + 2)

3 + e

x 3(4x

3 + 2)

2 12x

2 = e

x (4x

3 + 2)

2 [(4x

3 + 2) + 36x

2] =

= ex (4x

3 + 2)

2 (4x

3 + 36x

2 + 2)

d) 3

53

2ln

(5 2 )

xy

x

=

3

3

1 2ln

5 (5 2 )

x

x =

3 31ln 2 ln(5 2 )

5x x =

= 1

5[ln 2 + ln x

3 – 3ln(5 – 2x)] =

1

5[ln 2 + 3ln x – 3ln(5 – 2x)]

y ' = 1 3 2

0 35 5 2x x

=

1 3 6

5 5 2x x


a) f(x) = 2 4

2

( 5)

3

x

x

b) g(x) = e

7x(x – 5x

2)3 c) h(x) =

2·ln( 1)

2

x x

x

f(x) = 2 4

2

( 5)

3

x

x

f '(x) = 22

42232

)3(

)2()5()3(2)5(4

x

xxxxx

=



= 22

42232

)3(

)5(2)3()5(8

x

xxxxx

=

22

2232

)3(

)]5(2)3(8[)5(

x

xxxxx

=

= 22

3332

)3(

]102824[)5(

x

xxxxx

=

22

332

)3(

)146()5(

x

xxx

g(x) = e7x

(x – 5x2)3

3

g'(x) = 7e7x

(x – 5x2)3 + e

7x 3(x – 5x

2)2(1 – 10x) =

= e7x

(x – 5x2)2 [7(x – 5x

2) + 3(1 – 10x)] = e

7x(x – 5x

2)2 [7x – 35x

2 + 3 – 30x] =

= e7x

(x – 5x2)2 (– 35x

2 – 23x + 3)

h(x) = 2·ln( 1)

2

x x

x

h'(x) = 2

2

2

2

)2(

)1ln()2(1

2)1ln(

x

xxxx

xxx

=

= 2

2

2

222

)2(

)1ln()2(1

2)1ln()1(

x

xxxx

xxx

=

= 2

2

22222

)2(

1

)1ln()1()2(2)1ln()2)(1(

x

x

xxxxxxxx

=

= 22

22222

)2)(1(

)1ln()1()2(2)1ln()2)(1(

xx

xxxxxxxx


a) f(x) = e3x

·ln(2x – 5) b) g(x) = 2

2

3

1

x

x c) h(x) = (3x

2 + 5x – 1)

6 + x

2 – ln x

a) f(x) = e3x

·ln(2x – 5)

f '(x) = 3e3x

ln(2x – 5) + e3x

52

2

x =

52

2)52ln(33

xxe x

b) g(x) = 2

2

3

1

x

x

g'(x) = 22

222

)1(

3·2)1)(3(ln3·2

x

xx xx

c) h(x) = (3x2 + 5x – 1)

6 + x

2 – ln x

h'(x) = 6(3x2 + 5x – 1)

5(6x + 5) + 2x –

x

1

25) Derivar: g(x) = (x2 + 1)

2 · ln(e

3x + 4)

g '(x) = 2(x2 + 1)2x ln(e

3x + 4) + (x

2 + 1)

2

4

33

3

x

x

e

e =



= (4x3 + 4x) ln(e

3x + 4) +

4

)1(33

223

x

x

e

xe


a) 2

3 2

3 1

(2 3)

xy

x

b) 7 3 42 ln( 1)xy x

a) y ' = 43

23223

)32(

6)32(2)13()32(6

x

xxxxx =

= 43

2233

)32(

]12)13()32(6)[32(

x

xxxxx =

33

244

)32(

12361812

x

xxxx =

= 33

24

)32(

181224

x

xxx

b) y ' = 7·27x – 3

ln 2 ln(x4 +1) + 2

7x – 3

1

44

3

x

x = 2

7x – 3

1

4)1ln(2ln7

4

34

x

xx


a) 2

3 2

5 1

(2 3)

xy

x

b) 7 3 43 ln( 1)xy x

a) y ' = 43

23223

)32(

6)32(2)15()32(10

x

xxxxx =

= 43

2233

)32(

]12)15()32(10)[32(

x

xxxxx =

33

244

)32(

12603020

x

xxxx =

= 33

24

)32(

301240

x

xxx

b) y ' = 7·37x – 3

ln 3 ln(x4 +1) + 3

7x – 3

1

44

3

x

x = 3

7x – 3

1

4)1ln(3ln7

4

34

x

xx


a) f(x) = 5

3 2

2

( 1)

x

x b) g(x) = 4x

3 ln(3x + 1)

a) f(x) = 23

5

)1(

2

x

x

f '(x) = 43

235235

)1(

3)1(22)1)(2(ln2·5

x

xxx xx

=

= 43

2353

)1(

]6)1)(2(ln5[2)1(

x

xxx x

= 33

235

)1(

]62ln)55[(2

x

xxx

b) g(x) = 4x3 ln(3x + 1) g'(x) = 12x

2 ln(3x + 1) +

13

34 3

x

x =

= 12x2 ln(3x + 1) +

13

12 3

x

x




a) f(x) = 3

ln2 1

x

x b) g(x) =

5

4 2

2

( 1)

x

x

a) Simplificamos antes de derivar: f(x) = 12

ln3 x

x = ln x – ln (2x

3 + 1). Ahora

derivamos: f '(x) = 12

613

2

x

x

x =

xx

xx

4

33

2

612 =

xx

x

4

3

2

14

b) g(x) = 5

4 2

2

( 1)

x

x g'(x) =

44

345245

)1(

4)1(22)1)(2(ln2·5

x

xxx xx

=

= 44

3445

)1(

]4·2)1)(2(ln5)[1(2

x

xxxx

= 34

345

)1(

]82ln)55[(2

x

xxx


a) f(x) = 2 1

2

3

(2 1)

x

x

b) g(x) = (4x

3 – 6x)

2 ln x

a) f '(x) =4

12212

)12(

2)12(23)12)(3(ln3·2

x

xx xx

=4

12

)12(

]4)12)(3·(ln2)[12(3

x

xxx

=

= 3

12

)12(

]43ln)24[(3

x

xx

b) g '(x) = x

xxxxxx1

)64(ln)612)(64(2 2323 =

=

x

xxxxxx

64ln)612(2)64(

323 = 64ln)1224()64( 223 xxxxx


f(x) = 3

ln4 2

x

x

g(x) =

2

3

3

2 3

x

x

Simplificamos la primera función antes de derivarla:

f(x) = 3

ln4 2

x

x

= ln x – ln (4x

3 – 2) f '(x) =

24

1213

2

x

x

x =

xx

xx

24

12244

33

=

= xx

x

24

284

3

=

)2(2

)14(24

3

xx

x

=

xx

x

4

3

2

14

g(x) = 2

3

3

2 3

x

x g'(x) =

23

2232

)32(

63)32)(3(ln3·2

x

xx xx

= 23

232

)32(

]63ln)64[(3

x

xxx


a) f(x) = 2

3 2

1

( 1)

x

x

f '(x) = 43

23223

)1(

3)1(2)1()1(2

x

xxxxx=

43

2233

)1(

]6)1()1(2)[1(

x

xxxxx=



= 33

424

)1(

6622

x

xxxx =

33

24

)1(

264

x

xxx

b) g(x) = (4x2 – 3)e

–3x²

g'(x) = 8x e–3x²

– 6x e–3x²

(4x2 – 3) = e

–3x²(8x – 24x

3 + 18x) = e

–3x²(–24x

3 +26x)

c) h(x) = 233 5 1x

h'(x) = 3 22 )15(3

103

x

x =

3 22 )15(

10

x

x =

3 2

3 2

3 22 15

15

)15(

10

x

x

x

x =

3 32

3 2

)15(

15 10

x

xx=

= 15

15 102

3 2

x

xx

d) j(x) = log(x2 + x + 1)

j '(x) = 1

12

10ln

12

xx

x


a) y = 2

4 2

(3 )

x

x

y ' = 4

2

)3(

))3(2)(24()3(2

x

xxx

=

4)3(

]2)24()3(2)[3(

x

xxx

=

= 3)3(

4826

x

xx

=

3)3(

22

x

x

b) g(x) = (3 – 7x2)e

–3x²

g'(x) = –14x e–3x²

– 6x e–3x²

(3 – 7x2) = e

–3x² (–14x –18x + 42x

3) =

= e–3x²

(42x3 – 32x)

c) h(x) = 25x

+ 2

1

x

h'(x) = 5·25x

ln(2) – 3

2

x

d) j(x) = log(x2 + 2x – 1)

j '(x) = 12

22

10ln

12

xx

x

34) Calcule y simplifique las derivadas de:

f x(x2 1) (3x

3 5)

3 g(x) =

3

ln(2 )x

x

e h(x) = log(3x

2 – x)

f x(x2 1) (3x

3 5)

3

f '(x) = 2x (3x3 + 5)

3 + (x

2 – 1) 3 (3x

3 + 5)

2 9x

2 =

= 2x (3x3 + 5)

3 + 27x

2 (x

2 – 1) (3x

3 + 5)

2 =

= (3x3 + 5)

2 [2x (3x

3 + 5) + 27x

2 (x

2 – 1)] =



= (3x3 + 5)

2 (6x

4 + 10x + 27x

4 – 27x

2) = (3x

3 + 5)

2 (33x

4 – 27x

2 + 10x)

g(x) = 3

ln(2 )x

x

e

g '(x) = 23

33

)(

3)2ln(2

2

x

xx

e

exex

= x

xx

e

x

exx

x

e

6

33 3)2ln(

= x

xx

e

x

xxee

6

33 )2ln(3

=

= x

xx

xe

xxee6

33 )2ln(3 =

x

x

xe

xxe6

3 ))2ln(31( =

xxxe

xx36

)2ln(31

=

xxe

xx3

)2ln(31

h(x) = log(3x2 – x)

h '(x) = xx

x

23

16

)10ln(

1


a) 31 3

( ) 5 2x

f x xx

.

f '(x) = 5)25(3)31(3 2

2

x

x

xx = 2

2)25(15

313

x

x

xx =

= 2

2)25(15

1

x

x

b) 2 2( ) 2 ln 2g x x x .

g '(x) = 2x ln(x2 + 2) + (x

2 + 2)

2

22 x

x = 2x ln(x

2 + 2) + 2x = 2x[1 + ln(x

2 + 2)]

c) 5 .3 x xh x e

h'(x) = 5·35x

ln 3 + ex


f(x) = 22x x

x

; g(x) = (x

2 + 1)

2 · ln(e

3x + 4)

f '(x) = 2

2 1)·2()22ln2(

x

xxx xx =

2

22 222ln2

x

xxx xx =

2

2 22ln2

x

xx xx

g'(x) = 2(x2 + 1)2x ln(e

3x + 4) + (x

2 + 1)

2

4

33

3

x

x

e

e =

= 4

)1(3)4ln()44(

3

22333

x

xx

e

xeexx


a) f(x) = 2

2

2

(1 )

x x

x



f '(x) = 2 2

4

( 4 1)(1 ) ( 2 )2(1 )( 1)

(1 )

x x x x x

x

=

= 2

4

(1 )[( 4 1)(1 ) 2( 2 )]

(1 )

x x x x x

x

=

2 2

3

4 4 1 4 2

(1 )

x x x x x

x

=

3

3 1

(1 )

x

x

b) g(x) = 31 3 22 (1 )x x

g'(x) = 3 31 2 3 2 1 3 22 ( 3 )(ln 2)(1 ) 2 2(1 )( 3 )x xx x x x =

= 31 2 3 32 ( 3 )(1 )[(ln 2)(1 ) 2]x x x x =

31 5 2 32 (3 3 )[(1 )ln 2 2]x x x x

Hay otras posibilidades de simplificación, procedentes de sacar factor común,

pero nos conformaremos con ésta.

c) h(x) = log (1 – x3)

h'(x) =

2

3

3 1

1 ln10

x

x


a) f(x) = (x3 + 1) e

7x

f '(x) = 3x2 e

7x + (x

3 + 1) 7 e

7x = e

7x (7x

3 + 3x

2 + 7)

b) g(x) = 3x ln(2x + 5)

g'(x) = 3x ln 3 ln(2x + 5) + 3

x

2

2 5x =

23 ln 3ln(2 5)

2 5

x xx

c) h(x) = 2

2

2

( 1)

x

x

h'(x) = 2 2

4

2 ( 1) ( 2)2( 1)

( 1)

x x x x

x

=

2

4

( 1)[2 ( 1) 2( 2)]

( 1)

x x x x

x

=

= 2 2

3

2 2 2 4

( 1)

x x x

x

=

3

2 4

( 1)

x

x

39) Calcule las siguientes derivadas, simplificando los resultados:

a) g'(3), siendo g(x) = 2xe3x –1

.

g '(x) = 2e3x – 1

+ 2x3e3x – 1

= e3x – 1

(2 + 6x) g'(3) = 20e8

b) f(x) = 2

3 2

1

( 1)

x

x

f '(x) = 3 2 2 3 2

3 4

2 ( 1) (1 )2( 1)3

( 1)

x x x x x

x

=

= 3 3 2 2

3 4

( 1)[ 2 ( 1) (1 )6 ]

( 1)

x x x x x

x

=

4 2 4

3 3

2 2 6 6

( 1)

x x x x

x

=

4 2

3 3

4 6 2

( 1)

x x x

x



c) h(x) = log(x2 + x + 1)

h'(x) = 2

2 1 1

1 ln10

x

x x


a) 31 3

( ) 5 2x

f x xx

.

f '(x) = 5)25(3)31(3 2

2

x

x

xx = 2

2)25(15

313

x

x

xx =

= 2

2)25(15

1

x

x

b) 2 2( ) 2 ln 2g x x x .

g '(x) = 2x ln(x2 + 2) + (x

2 + 2)

2

22 x

x = 2x ln(x

2 + 2) + 2x = 2x[1 + ln(x

2 + 2)]

c) 5 .3 x xh x e

h'(x) = 5·35x

ln 3 + ex

Matemáticas Cálculo de derivadas PROBLEMAS ... - e … resueltos de... · Tomamos ln antes de...

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Transcript of Matemáticas Cálculo de derivadas PROBLEMAS ... - e … resueltos de... · Tomamos ln antes de...